## Brieskorn resolutions via algebraic spaces

I’d like to discuss simultaneous resolutions of surfaces from a moduli-theoretic perspective, following Michael Artin’s paper on Brieskorn resolutions.

Artin’s approach to moduli begins with the most desirable aspect of a moduli space, its universal property. That is to say, define a functor the space should represent and then check some local criteria to establish it’s algebraicity. Among the first successes of this method was the Hilbert, Quot, and Picard schemes (see his Formal Moduli I paper) but later Artin would go on to show that modifications (e.g. contractions) could also be represented by algebraic spaces, even if they weren’t representable by schemes. Using this idea, he showed that the category of algebraic spaces algebraically reproduces the category of Moishezon spaces in complex-analytic geometry (see his Formal Moduli II paper).

Here the primary goal will be to prove another algebraic avatar of a result in complex analytic geometry:

**Theorem 1 (Brieskorn)** Let denote a flat family of algebraic surfaces over having only rational double points as singularities, then there is an analytic space and a finite surjective morphism such that admits a simultaneous resolution of singularities.

**(*)** To fix ideas, from here onwards let denote a flat, separated and finite-type morphism of schemes *over a field* whose singular locus is finite over and whose geometric fibers are all normal and purely two-dimensional.

**Definition 2** Let be as above, a proper morphism is a *resolution *if is smooth and on each fiber the induced morphism is a minimal resolution of singularities. Two resolutions of are isomorphic if they are isomorphic over .

Artin begins by defining a functor on the category of -schemes:

set of resolutions of upto isomorphism

If we want to understand the extent to which can be simultaneously resolved, we had better understand this functor. The goal is to represent by an algebro-geometric object since, by definition, its points correspond to resolutions of the relative surface . In particular, it has a desirable universal property: the surface can be simultaneously resolved over an -scheme if and only if admits a -point. In fact, the representability follows from:

**Theorem 3 (Artin) **Let be as in *, then is a locally quasi-separated algebraic space.

We leave the proof of this result to a future post and grant it for now. Here is the algebraic analogue of Brieskorn’s result:

**Theorem 4 (Artin) **Suppose is as in (*) and is a versal deformation of a single rational singularity on . Let be the structure morphism and suppose gets mapped to . Then the local ring is nonsingular and the canonical morphism

maps the domain onto an irreducible component of . In particular, if the latter is irreducible, then the morphism is finite and surjective.

**Remark: **Note that in order for the last sentence of the Theorem to apply, we need the Henselian local rings of the base scheme to be unibranch, a condition which is guaranteed if is normal. However, by hypothesis if denotes the category of Artin local rings with residue field for some , the map induced by , is formally smooth (this is just a reformation of versality). Thus, if is formally smooth over (i.e. if the deformations of are unobstructed, a situation that holds if has rational double or triple points) we may deduce that is unibranch at and hence that the associated maps on local rings are both finite and surjective, a direct analogue of Brieskorn’s analytic result.

In the next post we’ll explain the proof of Theorem 4. Next, we’ll give some elementary observations and then conclude with two interesting examples.

**Proposition 5: **Let be as in (*) above and suppose is the resolution functor with mapping to .

(a) If is a morphism, then there is a natural identification .

(b) The structure morphism induces a bijection on -valued points for any field .

(c) If is smooth, then the functor is the singleton functor i.e. the structure morphism is an isomorphism.

(d) If is smooth at a point , then the morphism is an isomorphism in an open neighborhood of .

(e) The induced map is finite for any .

(f) If is a finite open cover of such that is smooth for all , then the natural morphism defined by restriction is an isomorphism of algebraic spaces.

**Proof: **Both (a) and (c) are easy to see so we omit their proofs. For (b) recall that minimal resolutions of surfaces always exist and are unique upto unique isomorphism. As for (d), note that by the finiteness of , there is an open neighborhood of over which is smooth, combining this with (a) yields the result: .

For (e), note that we may replace with since the formation of is compatible with restriction, so we assume is the spectrum of a Henselian local ring. Now, let

be an affine etale neighborhood with one point lying above and where is an isomorphism (see [Sta, Tag 0BBP]). It is quasifinite over and finite-type, thus the underlying -algebra can be written as where is a finite and is not quasi-finite at any prime lying over (see [Sta, Tag 04GE (11)]). Thus, since there cannot be a point in lying over , there must be such a point in and we may consider the finite etale map . By [Sta, Tag 04GH], is a Henselian local ring which is a finite -algebra. To conclude, note that can be identified with .Lastly, for (f) we need to show that the natural morphism of functors

is an equivalence. Indeed, if two resolutions and become isomorphic over the open cover , say, via isomorphisms , then these morphisms patch because the are all isomorphisms over . This shows injectivity, surjectivity follows because we may patch together a collection of resolutions . Indeed, for the glueing datum we use the isomorphisms (which easily satisfies the cocycle condition). Therefore the resulting resolution of restricts to the point defined by the resolutions . **QED**

The extent to which a given relative surface can be resolved simultaneously is reflected in the geometry of the morphism . For example, the worst case scenario is when is the disjoint union of the points in :

**Example: **A necessary condition for the simultaneous resolution of a family of surfaces is that each of their minimal resolutions must have the same arithmetic genus. In particular, a degeneration of a smooth family of surfaces has a chance of being simultaneously resolved only if the singularities on the special fiber are rational. Consider for instance, a pencil consisting of smooth surfaces for all and which yields the cone over an elliptic curve at . The minimal resolution has an elliptic curve as its exceptional locus which means the arithmetic genus of the surface must change along the resolution. Since the original family is flat it means the smooth members of the family and the resolution have differing Euler characteristics and hence they cannot appear in a flat family together. Thus, a simultaneous resolution cannot exist in this case and so if we replace the base of the pencil with the spectrum of a DVR specializing to , we see that must be a disjoint union of points.

**Example: **It is worth noting that the resulting algebraic space need not be a scheme, even in simple cases. Let be the spectrum of a DVR and a degeneration of a smooth quartic surface in whose general fiber has and whose special fiber has a single node . We claim that cannot be simultaneously resolved in the category of schemes. If so, this also shows that cannot be represented by a scheme since the universal resolution is a projective morphism. Indeed, the dual of the line bundle associated to the exceptional curve is a relatively ample line bundle. Thus, if was a scheme, and hence would be as well, a contradiction.

Let be a resolution so that is an isomorphism and is minimal. Next, choose a hyperplane section so that it’s closure misses and is therefore isomorphic to its preimage in . It follows that . If was a scheme there would be an open affine subset meeting , the complement would also be a Cartier divisor which is flat over . Indeed, we know any generic point of has height one and so if , it must be the generic point of the because the dimension of its closure must be two. This would mean that contains the special fiber, a contradiction by our choice of and so is -flat. We now have for some positive integer . To see this, observe that is trivial over the generic fiber (for some positive integer ) so we get the desired isomorphism by upper-semicontinuity. But the linear equivalence of and implies or, in other words, that and are disjoint. This means the projective scheme must live entirely inside the open affine subset , a contradiction.

That’s it for now, in the next post I will explain the proof of Theorem 4.